Convergence of Geometric Series .... Sohrab, Proposition 2.3.8 .... ....

In summary, Peter is struggling with a part of the proof of Proposition 2.3.8 and needs help from someone more experienced. The proof appears to be referring to Part 2 of Exercise 2.2.7, which states that if 0 \lt b \lt 1, then \lim ( b^n ) = 0. Peter is not sure how to proceed from here. Someone more experienced may be able to help him.
  • #1
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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with an aspect of the proof of Proposition 2.3.8 ...

Proposition 2.3.8 and its proof read as follows:
View attachment 9048
In the above proof by Sohrab we read the following:

" ... ... Now, if \(\displaystyle \mid r \mid \lt 1\) then (by Exercise 2.2.7) \(\displaystyle \lim (r^{ n + 1}) = 0\) and \(\displaystyle ( \ast )\) follows at once ... ... "
Now it appears that Sohrab is referring to Part 2 of Exercise 2.2.7 (see below) that states:

" ... ... if \(\displaystyle 0 \lt b \lt 1\), then \(\displaystyle \lim ( b^n ) = 0\) ... ... "Now my problem is we are not given \(\displaystyle 0 \lt r \lt 1\) ... but instead we are given \(\displaystyle \mid r \mid \lt 1\) ... .. so my question is ... what do we do here ...?
Perhaps we put \(\displaystyle b = \mid r \mid\) ... but this allows us to conclude \(\displaystyle \lim ( \mid r \mid^n ) = 0\) ...... can we conclude from here that \(\displaystyle \lim (r^{ n + 1}) = 0\) ...... by arguing that \(\displaystyle \lim ( \mid r \mid^n ) = 0 = \lim ( \mid r^n \mid ) = \lim ( \mid r^{ n +1} \mid )\)... and further by arguing that if \(\displaystyle \lim \mid r^{ n +1} \mid = 0\) ...... then ...... \(\displaystyle \lim r^{ n +1} = 0\) ...
Can someone please comment on my reasoning to resolve the problem I had with Sohrab's assertion that appeal to Exercise 2.2.7 would lead to \(\displaystyle ( \ast )\) following at once ...
Help will be appreciated ...

Peter
========================================================================================
The above post refers to Exercise 2.2.7 ... so I am providing text of the same ... as follows:View attachment 9049
Hope that helps ...

Peter
 

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  • #2
A nice way to do this would be with cauchy sequences, but I don't know if you've gotten to them yet. The argument would be

$m \gt n \geq N$$\big \vert s_n - s_m\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(1 -r^{n+1} -(1 -r^{m+1})\big)\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(- r^{n+1} + r^{m+1}\big)\big \vert $
$ = \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\cdot \big \vert r^{m-n} - 1 \big \vert $
$ \lt 2 \cdot \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1} $

where the inequality follows from triangle inequality

select $\frac{\epsilon}{2}$
i.e. based on earlier work with the non-negative case, you know
$\big(\big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\big)$ tends to zero... so re-use this result
- - - -
as far as your suggested approach, I think it works.

You can also observe that
$\big \vert r\big \vert^n =\big \vert r^n \big \vert = \big \vert r^n - L \big \vert$
with $L =0$

since you know from that prior exercise 2.2.7 that for any $\epsilon \gt 0$
$\big \vert r\big \vert^n \lt \epsilon$
by selecting large enough $N$, then you also know by selecting that same $N$ that
$\big \vert r^n - L \big \vert \lt \epsilon$
which is your definition of a limit
 
  • #3
steep said:
A nice way to do this would be with cauchy sequences, but I don't know if you've gotten to them yet. The argument would be

$m \gt n \geq N$$\big \vert s_n - s_m\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(1 -r^{n+1} -(1 -r^{m+1})\big)\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(- r^{n+1} + r^{m+1}\big)\big \vert $
$ = \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\cdot \big \vert r^{m-n} - 1 \big \vert $
$ \lt 2 \cdot \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1} $

where the inequality follows from triangle inequality

select $\frac{\epsilon}{2}$
i.e. based on earlier work with the non-negative case, you know
$\big(\big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\big)$ tends to zero... so re-use this result
- - - -
as far as your suggested approach, I think it works.

You can also observe that
$\big \vert r\big \vert^n =\big \vert r^n \big \vert = \big \vert r^n - L \big \vert$
with $L =0$

since you know from that prior exercise 2.2.7 that for any $\epsilon \gt 0$
$\big \vert r\big \vert^n \lt \epsilon$
by selecting large enough $N$, then you also know by selecting that same $N$ that
$\big \vert r^n - L \big \vert \lt \epsilon$
which is your definition of a limit
Thanks steep ...

Most grateful for your help ...

Peter
 

FAQ: Convergence of Geometric Series .... Sohrab, Proposition 2.3.8 .... ....

What is the Convergence of Geometric Series?

The Convergence of Geometric Series is a mathematical concept that refers to the behavior of a series where each term is a constant multiple of the previous term. In this type of series, the terms either increase or decrease exponentially, resulting in a specific pattern of convergence or divergence.

Who is Sohrab and what is Proposition 2.3.8?

Sohrab is a mathematician who is known for his work in the field of geometric series and their convergence. Proposition 2.3.8 is a specific theorem in Sohrab's work that deals with the convergence of geometric series and provides a formula for calculating their sum.

How is the convergence of geometric series determined?

The convergence of geometric series is determined by the common ratio between consecutive terms. If the common ratio is less than 1, the series will converge. If the common ratio is greater than or equal to 1, the series will diverge.

What is the formula for calculating the sum of a convergent geometric series?

The formula for calculating the sum of a convergent geometric series is S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio between consecutive terms.

How is the convergence of geometric series useful in real-world applications?

The convergence of geometric series has many practical applications in fields such as finance, physics, and engineering. It is used to model exponential growth and decay, calculate compound interest, and analyze the behavior of electrical circuits. It is also used in the study of infinite series and their applications in computer science and statistics.

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